Projectile Review Worksheet

From TuHSPhysicsWiki
Revision as of 11:23, 4 November 2008 by PHerstedt (talk | contribs) (2. Question)
Jump to: navigation, search

Main Page > Solutions to Worksheets > Projectile Review Worksheet

About this page

All images uploaded for this page must start with the string "Projectile_Review_Worksheet_" so the image 16-38.jpg associated with this page should be uploaded as Projectile_Review_Worksheet_16-38.jpg. This way we can avoid conflicts in the image directory, and we can find images easily.
Table of Contents


Problems

1. Marlene jumps off the edge of a cliff and hits the water 1.5 seconds later, about 4.5 m from the base of the cliff. What height was the cliff? With what speed did she leave the edge?

So we fill in our H/V suvat:

H V

s = 4.5 m
u =
v =
a = 0
t = 1.5 s

s =
u = 0 (cliff)
v =
a = -9.81 m/s/s
t = 1.5 s

We are looking for vertical displacement (s) and horizontal velocity (u). We can solve for each variable separately. For vertical displacement, we can use the formula

s = ut + 1/2at2

s = (0 m/s)(1.5 s) + (.5)(-9.81 m/s2)(1.5 s)2

s = 11.03625 m = 11 m

For horizontal velocity, we can use the formula

u = s/t

u = (4.5 m) / (1.5 s)

u = 3 m/s


Table of Contents


2. Question

H V

s = 3.4 m
u =
v =
a = 0
t =

s = 21
u = 0 (cliff)
v =
a = -9.81 m/s/s
t =

The first thing we are going to solve for is time (t). We will do this using the vertical side of the table.

s = ut + 1/2at2
21 m = (0 m/s)(t) + (.5)(9.81 m/s2)(t2)
21 m = 0 + (4.905)(t2)
t = 2.0691 s = 2.07 s

With our newfound value of time, we can now solve for the horizontal velocity (u).

u = s/t
u = (3.4 m) / (2.07 s)
u = 1.6425 m/s = 1.64 m/s

Table of Contents


3. Question

Solution goes here

Table of Contents


4. Question

Solution goes here

Table of Contents


5. Question

Solution goes here

Table of Contents


6. Question

Solution goes here

Table of Contents